How do you solve $log_3 +log_3 (2x)=log_3 56$ ?

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Answer 1 · 2016-08-10T14:20:49

The operand is missing in the first term. If it is a, the answer is $28/a$ .

Assuming that the missing operand in the first term is a,

the equation is

$log_3 a + log 3 (2x) = log_3 56$

So, $log_3 (a X 2x)=log_3 56.$

And so, $2ax=56 and x =28/a$ .

How do you solve log_3 +log_3 (2x)=log_3 56log_3 +log_3 (2x)=log_3 56?